Proof that $\mathbb{R} \not\sim \mathbb{R}^{\mathbb{R}}$

Question

I have troubles understanding the proof that $\mathbb{R} \not\sim \mathbb{R}^{\mathbb{R}}$ (that is, $|\mathbb{R}| \neq |\mathbb{R}^{\mathbb{R}}|$) where $\mathbb{R}^{\mathbb{R}}$ is the set of all functions from $\mathbb{R}$ to $\mathbb{R}$. The proof we did in the class is very messy and not well structured so I can't even comprehend basic ideas. We used the fact that $\mathbb{R} \sim [0, 1]$.

This proposition proves that there are "biggers" infinities than others, which is quite surprising!

Thanks in advance.

There are at least as many functions $\mathbb{R}\to\mathbb{R}$ as there are functions $\mathbb{R}\to{0,1}$. Every function $\mathbb{R}\to{0,1}$ corresponds to a subset of $\mathbb{R}$ and vice versa. So the statement can be proven by showing that $\mathbb{R}$ has cardinality lower than its power set. And that is the well known Cantor's theorem. — freakish, Apr 21 '23 at 08:03
$|\mathbb{R}^{\mathbb{R}}|\ge |{0,1}^{\mathbb{R}}|=|\mathcal P(\Bbb R)|>|\mathbb{R}|$ by Cantor's theorem. — Anne Bauval, Apr 21 '23 at 08:04
Does this answer your question? Cardinality of the set of all real functions of real variable (it was the 2011 target of this 2018 duplicate: https://math.stackexchange.com/questions/3009704). — Anne Bauval, Apr 21 '23 at 08:13
How is $[0, 1]^{\mathbb{R}}$ power set of $\mathbb{R}$? Is this because for any $x \in \mathbb{R}$ there exist a function $\mathbb{R} \rightarrow [0, 1]$? — 1b3b, Apr 21 '23 at 08:14
https://math.stackexchange.com/questions/41006/how-to-show-equinumerosity-of-the-powerset-of-a-and-the-set-of-functions-from. But first, beware we both wrote ${0,1}$ (the pair), not $[0,1]$ (the interval). — Anne Bauval, Apr 21 '23 at 08:21
This post should be closed, as an exact duplicate of The set of real numbers and the set of Real valued functions are not similar (equinumerous). — Anne Bauval, Apr 21 '23 at 09:47
So, codomain of every $f \in \mathbb{R}^{\mathbb{R}}$ is a set of ordered pairs $(a, 0)$ and $(b, 1)$ for $a, b \in \mathbb{R}$, so we get this binary relation (0 = "don't include", 1 = "include") which in turn makes bijection between $\mathbb{R}^{\mathbb{R}}$ and $\mathcal{P}(\mathbb{R})$. Minimalistic and beautiful. — 1b3b, Apr 21 '23 at 10:54

score 3 · Accepted Answer · answered Apr 21 '23 at 08:47

3

Here is a direct proof that $|\mathbb{R}| < |\mathbb{R}^\mathbb{R}|$, more precisely, that there is no surjection $s : \mathbb{R} \to \mathbb{R}^\mathbb{R}$.

Consider any function $s : \mathbb{R} \to \mathbb{R}^\mathbb{R}$. We now define the function $d : \mathbb{R} \to \mathbb{R}$ by $d(x) = s(x)(x) + 1$, and claim that $d$ is not in the range of $d$. Assume otherwise. Then there is some $y \in \mathbb{R}$ with $s(y) = d$. But then we'd conclude that $d(y) = s(y)(y) + 1 = d(y) + 1$, contradiction.

answered Apr 21 '23 at 08:47

Arno

3,788

This answer can be deleted, as an exact duplicate of that one. – Anne Bauval Apr 21 '23 at 09:49
Thank you, I think this is the proof we have done, but not as structured (not even close). – 1b3b Apr 21 '23 at 10:04

Proof that $\mathbb{R} \not\sim \mathbb{R}^{\mathbb{R}}$

1 Answers1